Real Analysis: Is n-binary digit expansion of a number in [0,1] measurable?

Hello,

If I have a function where and is the n-binary digit expansion of , how can I show that is measurable?

The full question I'm looking at asks to show that is measurable.

I know that when we have a sequence of measurable functions, their sum will be measurable, and dividing a measurable function by n will still leave it measurable as well. Finally, when dealing with a sequence of measurable functions than their will also be measurable.

I have most of the puzzle figured out, but have never dealt with n-binary digits before. If I can just show that this expansion is measurable, everything else will fall into place.

My textbook tells me that if a sequence of functions converges, then it's limit will be measurable. Doesn't that apply in this case, since the n-binary digit expansion is converging to some point in [0,1]?

Thanks for the reply! So the preimages of the sequence of for any will be unions of measurable sets, thus making the function itself measurable? This explanation was much simpler than what I had in mind.

Since the only empty set occurs when =0, does this mean that the function is continuous as long as (almost everywhere)?

Wouldn't this also mean that is just 1/2? That seems to make sense, but I'm still very new to binary digit expansions and want to make certain.

Since the only empty set occurs when =0, does this mean that the function is continuous as long as (almost everywhere)?

I am not sure I understand. What do you mean that the empty set occurs? What function is continuous? The function is not continuous; it is equal to 1 and 0 on alternating segments. I believe has n - 1 points on [0, 1) where it is not continuous.